Draft 1 On the Lattice Structure of MIXMAX Random Number Generators Pierre L’Ecuyer Joint work with Paul Wambergue, Erwan Bourceret, and Marc-Antoine Savard MCQMC, Rennes, July 2018
Draft 2 MIXMAX Generators [Akopov, Savvidy, et al. (1991)] Output vector u i ∈ R k follow the recurrence u i = Au i − 1 mod 1 where u 0 ∈ [0 , 1) k and A is a k × k matrix such that: (1) det ( A ) = 1, so the linear transformation preserves volume; (2) the eigenvalues λ 1 , . . . , λ k of A are away from the unit circle. Then, for appropriate A , distance between trajectories that start from very close will diverge exponentially fast, as Θ( e hi ) in i steps, where h is the entropy: k � h = I [ | λ j | > 1] · log | λ j | . j =1 For almost all (irrational) u 0 , the trajectory is aperiodic and space-filling in [0 , 1) k .
Draft 3 Original MIXMAX proposal: 1 1 1 1 · · · 1 1 1 2 1 1 · · · 1 1 1 3 + d 2 1 · · · 1 1 A = A ( k , d ) = . 1 4 3 2 · · · 1 1 . ... . . 1 k k − 1 k − 2 · · · 3 2 Savvidy (2015) provides lower bounds on h that depend only on k for A = A ( k , d ), and shows that h is large, e.g., much larger than for the AWC and SWB generators of Marsaglia and Zaman (1991). However, all of this is for irrational state vectors u i , which cannot be implemented exactly.
Draft 4 The authors proposed to approximate the irrational recurrence by a rational one as follows. Take a large prime m and define x i = Ax i − 1 mod m in which x i = ( x i , 0 , . . . , x i , k − 1 ) t ∈ Z k m and u i = ( u i , 0 , . . . , u i , k − 1 ) = x i / m ∈ [0 , 1) k . This is a matrix LCG (Niederreiter 1986); gives a periodic recurrence! If det( A ) = 1, the maximal period is ( m k − 1) / ( m − 1), i.e., m − 1 times shorter than general matrix LCGs. Savvidy provides parameters ( k , d ) giving period ρ = ( m k − 1) / ( m − 1) for m = 2 61 − 1 with k from 8 to 3150. Even for k = 8, this already gives ρ ≈ 2 427 . Long enough period! He also provides an efficient implementation that uses only 2 k additions and one multiplication by d to compute the next vector x i at each step. Fast! MIXMAX became popular recently. Part of the ROOT library used at CERN, Geneva. In July 2016, I was at CERN, in a workshop devoted to the MIXMAX. Fred James was telling us: “We have a proof that this RNG passes all statistical tests; this is fantastic.” I was not fully convinced.
Draft 5 Other variants, with more parameters and flexibility MIXMAX-( m , k , d , c ) and MIXMAX-( m , k , d , c , b ): 1 1 1 1 · · · 1 1 1 2 1 1 · · · 1 1 1 c + 2 + d 2 1 · · · 1 1 A = A ( k , d , c ) = 1 2 c + 2 c + 2 2 · · · 1 1 1 3 c + 2 2 c + 2 c + 2 · · · 1 1 · · · 1 ( k − 2) c + 2 ( k − 3) c + 2 ( k − 4) c + 2 · · · c + 2 2 1 1 1 1 · · · 1 1 1 2 1 1 · · · 1 1 1 3 c + d + b 2 1 · · · 1 1 A = A ( k , d , c , b ) = 1 4 c + b 3 c + b 2 · · · 1 1 1 5 c + b 4 c + b 3 c + b · · · 1 1 · · · 1 kc + b ( k − 1) c + b ( k − 2) c + b · · · 3 c + b 2
Draft 6 Lattice structure of matrix LCGs Define successive output values of matrix LCGs as follows: = ( u 0 , u 1 , . . . , u k − 1 ) u 0 u 1 = ( u k , u k +1 , . . . , u 2 k − 1 ) u 2 = ( u 2 k , u 2 k +1 , . . . , u 3 k − 1 ) , etc. For any fixed s > 0, let Ψ s = { ( u 0 , u 1 , . . . , u s − 1 ) ∈ [0 , 1) s | x 0 ∈ Z k m } , the set of vectors of s successive values obtained from all possible initial states. We want this set to cover [0 , 1) s very evenly. More general: for any finite set I = { i 1 , . . . , i s } where 0 ≤ i 1 < · · · < i s , let Ψ s ( I ) = { ( u i 1 , . . . , u i s ) ∈ [0 , 1) s | x 0 ∈ Z k m } . Want this set to cover [0 , 1) s very evenly for all I in some large family, e.g., i s ≤ i ∗ and s ≤ s ∗ .
Draft 7 Lattice structure It is known that for each I , Ψ s ( I ) = L s ( I ) ∩ [0 , 1) s where L s ( I ) is a lattice in R s . It means that all the points are in equidistant hyperplanes at distance d s ( I ) = 1 /ℓ s ( I ) apart, where ℓ s ( I ) = length of shortest vector in dual lattice. Let d ∗ s ( n ) be the best (smallest) distance for a known lattice with density n = min( m k , m s ) in s dimensions. Normalized measure: S s ( I ) = d ∗ s ( n ) / d s ( I ) ∈ (0 , 1]. Want it close to 1. Example of worst-case figure of merit: For s ′ > 0, let M s ′ = I ⊆{ 1 ,..., s ′ } S s ( I ) . min If s ′ is very large, there are so many subsets that some of them must be bad. But for s ′ not too large, most MRGs and matrix LCGs have a good M s ′ .
Draft 8 Example: MRG of order k Multiple recursive generators (MRG): = ( a 1 x i − 1 + · · · + a k x i − k ) mod m , u i = x i / m . x i Equivalent to matrix LCG with k 0 1 · · · 0 . . ... . . . . A = 0 0 · · · 1 a k a k − 1 · · · a 1 For k = 7 and m = 2 63 − 52425 (a prime), we searched random parameters a = ( a 1 , . . . , a k ) that gave full period ρ = m k − 1 ≈ 2 441 , and computed M 10 for each. Within our limited time budget, we obtained 7041 such vectors and the values of M 10 ranged from 0.53858 (the best) to 0.14970 (the worst). Thus, values of M 10 smaller than say 0.1 are very rare.
Draft 9 How to compute ℓ s ( I ) ? A t / m ( A ν − 1 ) t / m ([ A ν ] r ) t / m I / m · · · 0 I 0 ... V = I 0 I and m · I 0 0 − A I . ... . W = . − A ν − 1 I − [ A ν ] r 0 I The rows of V form a basis of L s and the rows of W form a basis of the dual lattice L ∗ s . For a basis of L s ( I ), select the appropriate colums of V , and invert this basis modulo 1 to get a basis for L ∗ s ( I ). Then compute the length ℓ s ( I ) of a shortest nonzero vector in L ∗ s ( I ), using BB.
Draft 10 Lattice structure of MIXMAX ( m , k , d ) m · I 0 0 0 · · · − 1 − 1 − 1 − 1 · · · − 1 − 2 − 1 − 1 · · · − 1 − 3 − d − 2 − 1 · · · I 0 0 · · · − 1 − 4 − 3 − 1 · · · · · · W = − 1 − ( k − 1) − 2 . − k · · · − A 2 0 I . . . . ... . . − [ A ν ] r 0 I The other MIXMAX are similar, with a slightly different − A .
Draft 11 Any integer linear combination of rows of W belongs to the dual lattice L ∗ s . In particular, if k ≥ 2, for all the MIXMAX variants, we have w k +1 = ( − 1 , − 1 , − 1 , . . . , − 1 , 1 , 0 , . . . , 0) w k +2 = ( − 1 , − 2 , − 1 , . . . , − 1 , 0 , 1 , . . . , 0) ↑ coordinate k + 1 so w = w k +1 − w k +2 = (0 , 1 , 0 , . . . , 0 , 1 , − 1 , 0 , . . . , 0) ∈ L ∗ s for s ≥ k + 2 . ↑ coordinate k + 1 √ This vector has Euclidean length 3. Its presence implies that the successive output values satisfy ( u 1 + u k − u k +1 ) mod 1 = 0. That is, u 1 + u k − u k +1 = 0 or 1, because 0 ≤ u i < 1 for all i . √ So when I := { 1 , k , k + 1 } ⊆ I ′ , Ψ 3 ( I ′ ) is covered by only two planes, 1 / 3 apart.
Draft 12 An example with k = 8 MIXMAX-( m , k , d , c ) from Savvidy (2017), with m = 2 61 − 1, k = 8, d = 0, and c = 2 53 + 1. Then, whenever { 1 , 8 , 9 } = I ⊆ I ′ , the points of Ψ 3 ( I ′ ) are in only two parallel planes. This gives S s ( I ) = 6 . 69 × 10 − 19 for this I , and then M s ′ ≤ 6 . 69 × 10 − 19 for all s ′ ≥ k + 2. We applied the following empirical statistical tests to this generator. Collision test: Generate n points in s dimensions, using all coordinates of output vectors. Partition the cube [0 , 1) s in d s cubic cells. Count the number C of collisions (a point falling in an occupied cell). C should be approx. Poisson with mean λ = n 2 / (2 d s ). Repeat N times and count the total number of collisions, then the p -value. With s = 16, d = 8, n = 4 × 10 7 , N = 10, we get p ≈ 6 × 10 − 206 . We expect about 28 collisions and we get 314. If we take only the first three values of each block and s = 6, with d = 128, same n and N , we get p < 10 − 300 . We expect about 1,818 collisions and we get 116,218. Birthday spacings test: With s = 16, d = 16, n = 3 × 10 7 , N = 10, we get p < 10 − 300 .
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